Lec%2025_Full

Lec%2025_Full - IE 343 Engineering Economics Lecture 25:...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
IE 343 Engineering Economics Lecture 25: Chapter 6 - Comparison and Selection among Alternatives Instructor: Tian Ni Oct.26, 2011 IE 343 Fall 2011 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Case 3: useful lives > study period Solution: The strategy is to truncate the alternative at the end of the study period. We need to find the market value of the alternative at the end of the study period, and we use the imputed(implied) market value technique to do this. IE 343 Fall 2011 Case 3: useful lives > study period 2
Background image of page 2
Let T denote the study period ( T < useful life), and let MV T denote the market value at time T . Then, MV T = PW at the end of year T of remaining capital recovery(CR) amount + PW at the end of year T of original market (salvage) value at end of useful life. Formally, let I be the initial investment and S be the salvage value of a project with a useful life of U time periods where U>T . Let i %=MARR, MV T = PW T (MARR) CR + PW T (MARR) S =[ I (A/P, i %, U )- S (A/F, i %, U )](P/A, i %, U-T )+ S (P/F, i %, U-T ). IE 343 Fall 2011 Case 3: useful lives > study period 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Consider a project with a capital investment of $47,600, useful life of 9 years, annual expenses of $1,720, and market (salvage) value of $5,000 at the end of its useful life. What is the estimated market value of this project at the end of year 5? The MARR is 20%. Note : Useful life = 9 years, Study period = 5 years, so it’s case 3: Useful life > Study Period IE 343 Fall 2011 Example 6.8 B Investment 47,600 Annual expenses 1,720 Useful life(years) 9 Salvage value 5,000 4
Background image of page 4
The estimated Imputed Market Value at the end of year 5 is: Recall the formula for computing imputed MV MV T = PW T (MARR) CR + PW T (MARR) S =[ I (A/P, i %, U )- S (A/F, i %, U )](P/A, i %, U-T )+ S (P/F, i %, U-T ) U: Useful Life = 9, T: Study Period = 5 MV 5 = [47,600(A/P,20%,9)-5,000(A/F,20%,9)](P/A,20%,4) +5,000(P/F,20%,4) = $32,358. IE 343 Fall 2011 Example 6.8 – Solution 5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
IE 343 Fall 2011 Example 6.8 0 $47,600 A=$1,720 A A 1 2 3 4 5 6 7 8 9 A A ………………………. .…. . Study period = 5 years S=$5,000 MARR = 20% 6
Background image of page 6
IE 343 Fall 2011 Example 6.8 – Solution 0 $47,600 A=$1,720 A A 1 2 3 4 5 A Study period = 5 years MARR = 20% Alternative B MV5 = $32,361 7
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Which of the following alternatives should be selected if the study period is 5 years and the MARR is 20%? Alternative B is identical to the project in Example 6.8. The imputed market value of Alternative B at the end of year 5 was found to be $32,358. IE 343 Fall 2011 Example 6.9 A B Investment 33,200 47,600 Annual expenses 2,165 1,720 Useful life(years) 5 9 Salvage value 0 5,000 8
Background image of page 8
We can use the AW method to compare the alternatives: AW(20%) A = -33,200(A/P,20%,5)-2,165 = -$13,266. AW(20%)
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 58

Lec%2025_Full - IE 343 Engineering Economics Lecture 25:...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online